C#:为什么旋转不可逆矩阵并再次旋转使其可逆
C#: Why does rotating a non invertible matrix and back again make it invertible
考虑以下测试代码:
using System;
using System.Drawing;
using System.Drawing.Drawing2D;
public class Program
{
public static void Main()
{
var matrix = new Matrix(123, 24, 82, 16, 47, 30);
Console.WriteLine(matrix.IsInvertible);
Console.WriteLine(matrix.Elements[0] + ", " + matrix.Elements[1] + ", " + matrix.Elements[2] + ", " + matrix.Elements[3]+ ", " + matrix.Elements[4]+ ", " + matrix.Elements[5]);
matrix.Rotate(90);
Console.WriteLine(matrix.IsInvertible);
Console.WriteLine(matrix.Elements[0] + ", " + matrix.Elements[1] + ", " + matrix.Elements[2] + ", " + matrix.Elements[3]+ ", " + matrix.Elements[4]+ ", " + matrix.Elements[5]);
matrix.Rotate(-90);
Console.WriteLine(matrix.IsInvertible);
Console.WriteLine(matrix.Elements[0] + ", " + matrix.Elements[1] + ", " + matrix.Elements[2] + ", " + matrix.Elements[3]+ ", " + matrix.Elements[4]+ ", " + matrix.Elements[5]);
}
}
在我的机器上输出
IsInvertible: False
Elements: 123, 24, 82, 16, 47, 30
IsInvertible: True
Elements: 82, 16, -123, -24, 47, 30
IsInvertible: True
Elements: 123, 24, 82, 16, 47, 30
这个结果让我感到惊讶 - 我旋转了一个不可逆矩阵,然后反转了旋转,得到了完全相同的元素。但是为什么矩阵不再是不可逆的呢?这个怪癖的原因是什么?
答案是 "loss of precision"。
你提供的数组确实是不可逆的
但是,当您将其旋转 90 度并再次旋转时,舍入误差导致原始数字略有不同 - 足以使矩阵现在可逆。
如果您将 .ToString("r")
添加到每个 WriteLines 以打印完整数字,您将看到:
122.999985, 23.9999962, 82, 16, 47, 30
注意前两个数字是如何变化的。
另请注意,您可以像这样更简洁地打印结果:
Console.WriteLine(string.Join(", ", matrix.Elements.Select(n => n.ToString("r"))));
考虑以下测试代码:
using System;
using System.Drawing;
using System.Drawing.Drawing2D;
public class Program
{
public static void Main()
{
var matrix = new Matrix(123, 24, 82, 16, 47, 30);
Console.WriteLine(matrix.IsInvertible);
Console.WriteLine(matrix.Elements[0] + ", " + matrix.Elements[1] + ", " + matrix.Elements[2] + ", " + matrix.Elements[3]+ ", " + matrix.Elements[4]+ ", " + matrix.Elements[5]);
matrix.Rotate(90);
Console.WriteLine(matrix.IsInvertible);
Console.WriteLine(matrix.Elements[0] + ", " + matrix.Elements[1] + ", " + matrix.Elements[2] + ", " + matrix.Elements[3]+ ", " + matrix.Elements[4]+ ", " + matrix.Elements[5]);
matrix.Rotate(-90);
Console.WriteLine(matrix.IsInvertible);
Console.WriteLine(matrix.Elements[0] + ", " + matrix.Elements[1] + ", " + matrix.Elements[2] + ", " + matrix.Elements[3]+ ", " + matrix.Elements[4]+ ", " + matrix.Elements[5]);
}
}
在我的机器上输出
IsInvertible: False
Elements: 123, 24, 82, 16, 47, 30
IsInvertible: True
Elements: 82, 16, -123, -24, 47, 30
IsInvertible: True
Elements: 123, 24, 82, 16, 47, 30
这个结果让我感到惊讶 - 我旋转了一个不可逆矩阵,然后反转了旋转,得到了完全相同的元素。但是为什么矩阵不再是不可逆的呢?这个怪癖的原因是什么?
答案是 "loss of precision"。
你提供的数组确实是不可逆的
但是,当您将其旋转 90 度并再次旋转时,舍入误差导致原始数字略有不同 - 足以使矩阵现在可逆。
如果您将 .ToString("r")
添加到每个 WriteLines 以打印完整数字,您将看到:
122.999985, 23.9999962, 82, 16, 47, 30
注意前两个数字是如何变化的。
另请注意,您可以像这样更简洁地打印结果:
Console.WriteLine(string.Join(", ", matrix.Elements.Select(n => n.ToString("r"))));